Problem 44 - Precalculus: Mathematical for Calculus

Question

The sum of two numbers is twice their difference. The larger number is 6 more than twice the smaller. Find the numbers.

Step-by-Step Solution

Answer

The numbers are 6 and 18.

Step 1: Define Variables

Let's assign variables to the unknowns. Let \( x \) be the smaller number and \( y \) be the larger number. We have two numbers to find.

Step 2: Translate the First Condition

The problem states that the sum of the two numbers is twice their difference. Mathematically, this can be represented as: \( x + y = 2(y - x) \).

Step 3: Simplify the First Equation

Simplify the equation from Step 2:\[x + y = 2y - 2x \ x + 2x + y = 2y \ 3x = y\]This represents our first equation.

Step 4: Translate the Second Condition

The problem also states that the larger number is 6 more than twice the smaller. This can be represented as: \( y = 2x + 6 \).

Step 5: Solve the System of Equations

We now have a system of two equations:1. \( y = 3x \)2. \( y = 2x + 6 \)Set them equal to find \( x \):\[ 3x = 2x + 6 \]Solve for \( x \):\[x = 6\]

Step 6: Find the Larger Number

Substitute \( x = 6 \) into either equation to find \( y \). Using \( y = 3x \):\[y = 3(6) = 18\]Thus, the larger number is \( y = 18 \).

Key Concepts

System of Equations

In algebra, a system of equations consists of two or more equations with the same set of unknowns. The goal is to find the values of these unknowns that satisfy all equations in the system. In this exercise, we are dealing with two equations derived from the problem statement.

To break it down:

• The first equation is derived from the condition that the sum of two numbers is twice their difference: \( x + y = 2(y - x) \), which simplifies to \( y = 3x \).
• The second equation comes from the condition that the larger number is 6 more than twice the smaller: \( y = 2x + 6 \).

Both equations represent lines on a graph, and the solution to the system is the point where these lines intersect. Systems of equations can often be solved through methods like substitution or elimination, allowing us to find the point of intersection that satisfies both conditions simultaneously.

Variable Substitution

Variable substitution is a powerful algebraic technique used to solve systems of equations. By replacing one variable with an expression involving the other, you can reduce the number of unknowns and solve the equations.

Here's how it works in this problem:

• First, we express one variable in terms of the other using one of the equations. From the equation \( y = 3x \), we already have \( y \) expressed in terms of \( x \). This makes it easier to substitute \( y \) in the other equation.
• Next, substitute this expression into the other equation \( y = 2x + 6 \). By setting \( 3x = 2x + 6 \), we are left with a single-variable equation.
• Solving \( 3x = 2x + 6 \) gives us the value of \( x \), simplifying our system significantly.

This technique is very useful when dealing with linear equations, as it often leads to straightforward solutions.

Problem-Solving Steps

Problem-solving in algebra starts with understanding the problem statement and translating it into mathematical expressions. Let's walk through the steps used in this exercise:

1. **Define Variables**: Assign symbols to unknown quantities. In this case, \( x \) for the smaller number and \( y \) for the larger.
2. **Translate Conditions**: Convert each condition into an equation. The first condition translates to \( x + y = 2(y - x) \), and the second to \( y = 2x + 6 \).
3. **Simplify Equations**: Rearrange and simplify the equations if necessary, as done when simplifying \( x + y = 2(y - x) \) to \( y = 3x \).
4. **Solve the System**: Use methods like substitution to find the values of the variables. Here, substituting and equating gives \( x = 6 \).
5. **Verify the Solution**: Once you have a solution, always substitute back to check if it satisfies all original equations. Substitute \( x = 6 \) into \( y = 3x \) and \( y = 2x + 6 \) to verify \( y = 18 \).Each step is crucial, providing a structured and organized way to approach algebra problems effectively.